What is the general solution for sec(x)sin(x)+cos(x)=sec(x) ?

The general solution for sec(x)sin(x)+cos(x)=sec(x) is x=2pin,x=pi+2pin

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Popular Trigonometry >

sec(x)sin(x)+cos(x)=sec(x)

Solution

sec (x) sin (x) + cos (x) = sec (x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sec (x) sin (x) + cos (x) = sec (x)

Subtract

sec (x)

from both sides

sec (x) sin (x) + cos (x) - sec (x) = 0

Express with sin, cos

\frac{1}{cos ( x )} sin (x) + cos (x) - \frac{1}{cos ( x )} = 0

Simplify

\frac{1}{cos ( x )} sin (x) + cos (x) - \frac{1}{cos ( x )} : \frac{sin ( x ) - 1 + cos ^{2} ( x )}{cos ( x )}

\frac{1}{cos ( x )} sin (x) + cos (x) - \frac{1}{cos ( x )}

\frac{1}{cos ( x )} sin (x) = \frac{sin ( x )}{cos ( x )}

\frac{1}{cos ( x )} sin (x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot sin ( x )}{cos ( x )}

Multiply:

1 \cdot sin (x) = sin (x)

= \frac{sin ( x )}{cos ( x )}

= \frac{sin ( x )}{cos ( x )} + cos (x) - \frac{1}{cos ( x )}

Combine the fractions

\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )} : \frac{sin ( x ) - 1}{cos ( x )}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( x ) - 1}{cos ( x )}

= \frac{sin ( x ) - 1}{cos ( x )} + cos (x)

Convert element to fraction:

cos (x) = \frac{cos ( x ) cos ( x )}{cos ( x )}

= \frac{sin ( x ) - 1}{cos ( x )} + \frac{cos ( x ) cos ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( x ) - 1 + cos ( x ) cos ( x )}{cos ( x )}

sin (x) - 1 + cos (x) cos (x) = sin (x) - 1 + cos^{2} (x)

sin (x) - 1 + cos (x) cos (x)

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

= sin (x) - 1 + cos^{2} (x)

= \frac{sin ( x ) - 1 + cos ^{2} ( x )}{cos ( x )}

\frac{sin ( x ) - 1 + cos ^{2} ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (x) - 1 + cos^{2} (x) = 0

Subtract

cos^{2} (x)

from both sides

sin (x) - 1 = - cos^{2} (x)

Square both sides

(sin (x) - 1)^{2} = (- cos^{2} (x))^{2}

Subtract

(- cos^{2} (x))^{2}

from both sides

(sin (x) - 1)^{2} - cos^{4} (x) = 0

Factor

(sin (x) - 1)^{2} - cos^{4} (x) : (sin (x) - 1 + cos^{2} (x)) (sin (x) - 1 - cos^{2} (x))

(sin (x) - 1)^{2} - cos^{4} (x)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

cos^{4} (x) = (cos^{2} (x))^{2}

= (sin (x) - 1)^{2} - (cos^{2} (x))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(sin (x) - 1)^{2} - (cos^{2} (x))^{2} = ((sin (x) - 1) + cos^{2} (x)) ((sin (x) - 1) - cos^{2} (x))

= ((sin (x) - 1) + cos^{2} (x)) ((sin (x) - 1) - cos^{2} (x))

Refine

= (cos^{2} (x) + sin (x) - 1) (sin (x) - cos^{2} (x) - 1)

(sin (x) - 1 + cos^{2} (x)) (sin (x) - 1 - cos^{2} (x)) = 0

Solving each part separately

sin (x) - 1 + cos^{2} (x) = 0 or sin (x) - 1 - cos^{2} (x) = 0

sin (x) - 1 + cos^{2} (x) = 0 : x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn

sin (x) - 1 + cos^{2} (x) = 0

Rewrite using trig identities

- 1 + cos^{2} (x) + sin (x)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= sin (x) - sin^{2} (x)

sin (x) - sin^{2} (x) = 0

Solve by substitution

sin (x) - sin^{2} (x) = 0

Let:

sin (x) = u

u - u^{2} = 0

u - u^{2} = 0 : u = 0, u = 1

u - u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- u^{2} + u = 0

Solve with the quadratic formula

- u^{2} + u = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 0 = 1

1^{2} - 4 (- 1) \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 1}{2 ( - 1 )}, u_{2} = \frac{- 1 - 1}{2 ( - 1 )}

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

\frac{- 1 + 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 1}{- 2 \cdot 1}

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 1 - 1}{2 ( - 1 )} : 1

\frac{- 1 - 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 1}{- 2 \cdot 1}

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 0, u = 1

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = 1

sin (x) = 0, sin (x) = 1

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn

sin (x) - 1 - cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn

sin (x) - 1 - cos^{2} (x) = 0

Rewrite using trig identities

- 1 - cos^{2} (x) + sin (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 1 - (1 - sin^{2} (x)) + sin (x)

Simplify

- 1 - (1 - sin^{2} (x)) + sin (x) : sin^{2} (x) + sin (x) - 2

- 1 - (1 - sin^{2} (x)) + sin (x)

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

- (1 - sin^{2} (x))

Distribute parentheses

= - (1) - (- sin^{2} (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + sin^{2} (x)

= - 1 - 1 + sin^{2} (x) + sin (x)

Subtract the numbers:

- 1 - 1 = - 2

= sin^{2} (x) + sin (x) - 2

= sin^{2} (x) + sin (x) - 2

- 2 + sin (x) + sin^{2} (x) = 0

Solve by substitution

- 2 + sin (x) + sin^{2} (x) = 0

Let:

sin (x) = u

- 2 + u + u^{2} = 0

- 2 + u + u^{2} = 0 : u = 1, u = - 2

- 2 + u + u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} + u - 2 = 0

Solve with the quadratic formula

u^{2} + u - 2 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

1^{2} - 4 \cdot 1 \cdot (- 2)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 \cdot 1}, u_{2} = \frac{- 1 - 3}{2 \cdot 1}

u = \frac{- 1 + 3}{2 \cdot 1} : 1

\frac{- 1 + 3}{2 \cdot 1}

Add/Subtract the numbers:

- 1 + 3 = 2

= \frac{2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 1 - 3}{2 \cdot 1} : - 2

\frac{- 1 - 3}{2 \cdot 1}

Subtract the numbers:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = 1, u = - 2

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - 2

sin (x) = 1, sin (x) = - 2

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sec (x) sin (x) + cos (x) = sec (x)

Remove the ones that don't agree with the equation.

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

sec (x) sin (x) + cos (x) = sec (x)

plug in

x = 2 π 1

sec (2 π 1) sin (2 π 1) + cos (2 π 1) = sec (2 π 1)

Refine

1 = 1

\Rightarrow True

Check the solution

π + 2 πn :

True

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

sec (x) sin (x) + cos (x) = sec (x)

plug in

x = π + 2 π 1

sec (π + 2 π 1) sin (π + 2 π 1) + cos (π + 2 π 1) = sec (π + 2 π 1)

Refine

- 1 = - 1

\Rightarrow True

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

sec (x) sin (x) + cos (x) = sec (x)

plug in

x = \frac{π}{2} + 2 π 1

sec (\frac{π}{2} + 2 π 1) sin (\frac{π}{2} + 2 π 1) + cos (\frac{π}{2} + 2 π 1) = sec (\frac{π}{2} + 2 π 1)

Refine

\infty = \infty

\Rightarrow True

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = 2 πn, x = π + 2 πn

Graph

Popular Examples

sin(x)+4csc(x)+5=0,0<= x<= 2pi

Frequently Asked Questions (FAQ)

What is the general solution for sec(x)sin(x)+cos(x)=sec(x) ?
The general solution for sec(x)sin(x)+cos(x)=sec(x) is x=2pin,x=pi+2pin